I am looking for a possibility to write the following sum as an integral, here $\Delta s$ is a partition of $[0,1]$ in $N$ pieces $$ \sum_{i=0}^{N-1} \big( f((i+1)\Delta s)-f(i\Delta s) \big)^2 =\int_0^1 \dots $$ And I am utterly confused, I had in my memory that probability theory does this all the time when using definitions of continuous prob. spaces for countable or finite ones. I don't find anything.
However, everything I wrote done was useless. Thanks for any help
$$\int_{-\infty}^\infty (f(x) - f(x - \Delta s))^2\sum_{i=1}^N\delta(x - i\Delta s)\, dx$$ is about as well as I can see to do it. You need your integral to cover both sides of any point where you are using the delta function, so $\int_0^1$ isn't going to cut it. But since $\delta(x)$ is $0$ away from $x = 0$, it doesn't matter how much extra range you throw in. You could also use $f(x + \Delta s) - f(x)$ by adjusting the summation to $\sum_{i = 0}^{N-1}$.